Question
Consider a circuit consisting of a battery with an emf $\varepsilon$ and an internal resistance of $r$ connected in series with a resistor $R$ and a capacitor $C$. Show that the total energy supplied by the battery while charging the battery is equal to $\varepsilon^{2} C$.
Step 1
In this case, the total resistance includes the internal resistance $r$ of the battery and the external resistance $R$, so $R = r + R$. Therefore, the equation for the current becomes $I(t) = \frac{\varepsilon}{r+R} e^{-t/(r+R)C}$. Show more…
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